Section 6.4 (p.336)

1. Follows Example 1. We have a = 5, b = 10, n = 20 and S x_i = 1. Therefore the Bayes estimator = (5+1) / (5+10+20) = 6/35.

2. (a). Let y be the number of defective items in the sample. The posterior distribution of q is Beta with parameters a = 5+y and b = 10+20-y = 30-y. Since the mean squared error of the Bayes estimate is the variance of the posterior distribution. Therefore we want to find y such that it maximizes the variance V = (5+y)(30-y) / (35)² (36) (see sec. 5.10).

Let dV/dy = 0, we have y = 12.5. But y has to be an integer, so y is either 12 or 13. Substituting these values into (5+y)(30-y) gives the same result. So 12 and 13 are both good answers.

(b). Since (5+y)(30-y) is a quadratic function of y and the coefficient of y² is negative, its minimum value over the interval 0 £ y £ 20 will be attained at one of the endpoints of the interval. It can be found that the value for y = 0 is smaller than the value for y = 20. Therefore the answer is 0.

8. Once again the mean squared error of the Bayes estimator of q is the variance of the posterior distribution of q. Follow Theorem 3. of section 6.3, we have s = 2 and n = 1. The posterior variance is . Let £ 0.01 and solve it for n, we get n ³ 396.

12.

Therefore, we have

That is, the posterior distribution of q is also a Pareto distribution, with parameters new a = a + n and new x₀ = max (x₀, x₁, …, x_n).

Using the squared error loss function, the Bayes estimator is the mean of the posterior distribution. So we have to compute the mean of a Pareto distribution. Suppose X has a Pareto distribution with a and x₀, then

Therefore, the Bayes estimator is .